1. Mathematics Revision Guide
Common Entrance
Level 1 and 2

2. Menu
Level 1 and 2 Topics
Click here.
This revision list contains revision for both levels.
Any topics in red are for level 2 only.

3. Click here to enter Number Revision Main menu
4 rules number and decimals
Arithmetic problems
Converting measures: Litres to ml and Grams to kg
Reading scales
Converting between % fractions decimals
Fractions of amounts
% problems % profit, profit as a %
BIDMAS
Writing numbers as product of prime factors using indices + use to find square number
Square and Square root
Unitary method
Distance, Time, Speed.
Ratio
Proportion
Using a calculator
Significant figures
Decimal places
Negative numbers
Subtraction and addition of simple fractions
x ÷ simple fractions
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Number Revision
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4. Click here to enter
Algebra Revision
‘Algebra simplifying expressions + - x ÷
Forming algebraic expressions
X brackets including x by negative number (level2).
Factorising expressions
Substitution including negative numbers ( level2)
Drawing line graphs using substitution and coordinates
Think of number’ inverse method
Solving equations
Sequences, term to term rule
Number patterns investigation
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5. Click here to enter
Shape and Space
Revision
Equations of lines and coordinates
Transformations: reflections rotations, translations. y=x
Enlargements
Angles
Exterior and interior angles of regular polygons
Bearings
Line Symmetry and Rotational Symmetry
Naming shapes
Construction
Nets of solids
Area and perimeter of rectangles and composite shapes
Area of triangle
Volume of cuboids
Volume, link between litres and cm³ (level 2)
Circumference of Circles
Area of Circles
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Data Handling
+ Graphs
Revision
Pie charts
Probability and combinations
Bar charts
range, mode, median, mean
Scatter Graphs
Conversion Graphs
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7. Topics
Level 1 and 2
Algebra
Shape and Space
Data Handling
+ Graphs
Number

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14. 2a + 3a + a = 6a 3c – c = 2c 2t -5t = -3t 3p x 4 = 12p t x t = t ²
Algebra
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Expressions
2a + 3a + a
= 6a
3c – c
= 2c
2t -5t
= -3t
3p x 4
= 12p
t x t
= t ²
t x t x t
= t ³
2t x 3t
= 6t ²
20a ÷ 4
= 5a
15c ÷ 3c
= 5

15. Carol has 2 less than Alan. Carol has m-2
Algebra
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Forming Expressions
Alan has m sweets
Bob has 3 more than Alan.
Bob has m+3
Carol has 2 less than Alan.
Carol has m-2
Diana has twice as much as Alan.
Diana has m
Edward has three times as much as Alan.
Edward has 3m
Altogether they have
m + m+3 + m-2 + 2m + 3m
= 8m + 1

16. 3( a + b) 2( a + b) 3 lots of (a+b) 2 lots of (a+b) a+b + a+b + a+b
Algebra
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Multiplying Brackets
3( a + b)
3 lots of (a+b)
a+b + a+b + a+b
= 3a + 3b
2( a + b)
2 lots of (a+b)
a+b + a+b
= 2a + 2b
4 + 3( 2a + 1)
4 plus 3 lots of 2a+1
4 + 2a a a+1
= 4 + 6a + 3
= 6a + 7
2( 3a + 2b)
2 lots of (3a+2b)
3a+2b + 3a+2b
= 6a + 4b

17. = 2(12a + 8b) or 4(6a + 4b) but they still have common factors
Algebra
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Factorising
2a + 4b
= 2(a + 2b)
6a + 4b
= 2(3a + 2b)
9a + 6b
= 3(3a + 2b)
24a + 16b
= 2(12a + 8b) or 4(6a + 4b) but they still have common factors
= 8(3a + 2b) is better
18a + 30
= 6(3a + 5)
10a + 15b
= 5(2a + 3b)

18. Solve 3w = 9 ÷3 ÷3 w = 3 x / 4 =12 x 4 x 4 x = 48 3y + 6 = 18 -6 -6
Algebra
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Equations
Find the value of………………Solve these equations……………
Solve
3w = 9
÷ ÷3
w = 3
x / 4 =12
means x divided by 4 =12
x x 4
x = 48
3y + 6 = 18
3y = 12
÷ ÷3
y = 4

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24. Adding and Subtracting
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Decimals
Adding and Subtracting
Sequences
Comparing Decimals

25. 0.34
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26. 0.51
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0.49 +
0.23 =

28. 0.72
0.49 +
0.23 =
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29. 0.31
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0.25 +
0.06 =

30. 0.85
0.25 +
0.6 =
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31. Sequences 1.0 1.2 0.8 Find the next 3 terms
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Sequences
Find the next 3 terms
A) 0.2, 0.4, 0.6, …. , …. , ….

32. 2.6
1.5
2.5
2.4
1.2
3.0
2.2
0.9
2.0
Find the next 3 terms
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0.5, 1.0, 1.5, …. , …. , …..
0 , 0.3 , 0.6, …. , …. , ….
D) , 1.8 , 2.0 , …. , …. , ….

33. 0.25
0.30
0.20
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E) , , , …. , …. , ….

34. Which is the largest number?
Are you sure?
Think of
strips and squares
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Which is the largest number? or

35. Which is the largest? A) 0.25 or 0.3 B) 0.5 or 0.71 C) 0.6 or 0.59
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Which is the largest?
A) or 0.3
B) or
C) or
D) or or

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Percentage Fractions

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Section 1
Building Percentages
10% 1%

39. 100% is equivalent to one whole30 35
3.5
45.5
0.4 60 70 7 91
0.8 15
17.5
1.75
22.75
0.2 3
3.5
0.35
4.55
0.04
Finding 10% is a useful starting point to finding percentages of amounts
100% is equivalent to one whole
10% means 10 out of every 100
10 % is one tenth
To find 10% find one tenth
÷ by 10
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Find 10% of
300
350 35
455 4
Find 20% of
300
350 35
455 4
Find 5% of
300
350 35
455 4
Find 1% of
300
350 35
455 4

40. Start with 10% and build these percentages
= 3.75
= 3.6
= 38.5
10%
10%
10%
10% 1% 5% = 21
0.5%
10% 5% 1%
Start with 10% and build these percentages
35%
35% of 60 =
15% of 24 =
15%
11% of 350 =
11%
1.5% of 250 =
1.5%
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41. Price after 10% rise
£20 + £2 = £22
Price after 5% rise
£22 + £1.10
= £23.10
Add this to the initial value. £
+ £ £
Find 28%
( £ X 3 ) – ( £1 500 X 2 )
= £
=£42 000
120 – ( )
= £ 102
30% - 2%
Problems
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1. If the value of a house increases by 28% from £ , find its new value.
2. If a coat is reduced 15% in the sale from £120, find its sale price.
3. If the price of a train ticket increases 10% one year and 5% the next from £20 , find the cost after the two increases.

42. Percentages expressed as Decimals
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Section 2
Percentages expressed as Decimals
Calculator work
0.01 is 1%
0.10 is 10%
0.08 is 8%
0.30 is 30%
is 2.5%
0.35 is 35%
is 0.5%
is 17.5%

43. Find these percentages6
160
2.75
39.6 6
3142.5
1.5
3.75
67500 6
6875
7.5
2.5
78750
Using a Calculator
Find these percentages
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Hint: Find 1% by dividing by 100.
Then multiply by the percentage you need to find.
2% of 300
3% of 250
8% of 2000
10% of 25
11% of 25
15% of 40
33% of 120
0.5% of 300
1.5% of 250
12.5% of 48
5.25% of
150% of
175% of
2.5% of

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Section 3
Finding the original value (100%) after a percentage increase or decrease
Given
120%
To find 1%
÷ 120
Then x 100
Given
80%
To find 1%
÷ 80
Then x 100

45. (10%) =
£12000
Problems to find the initial value after being given the value after an increase or decrease
The price of a new car increases by 10%.
The price after the increase is £
Find the price before the increase.
100%+10% = 110%
110% £
Find 1% ÷ 110 then x 100 =
Check the problem
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46. If a car loses 20% of its value and it is worth
£ 6 000
£ £1 200 (20%) = £4 800
80% £4 800
4800 ÷ 80 x 100 =
If a car loses 20% of its value and it is worth
£ , find its original value.
What percentage represents £4 800 ?
Now calculate the original value.
Remember to write down the numbers you use even on the calculator paper.
Check the problem
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Fractions of Amounts 47

48. Measures 1 1000 g = 1 kg How many grams is ½ kg ?
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Measures 1
1000 g = 1 kg
How many grams is ½ kg ?
How many grams is ¼ kg ?
How many grams is ¾ kg ? 48

49. Measures 2
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1 Kg
1000 g
250, 500, 750, , ,
¼ , ½ , ¾ , , ,
250 g
250 g
250 g
250 g 49

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Measures 3
100 cm = 1 m How many centimetres is ½ m ? How many centimetres is ¼ m ? How many centimetres is ¾ m ? 50

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Measures 4
1 m 100 cm 25 cm 25 cm 25 cm 25 cm 25, 50, 75, 100, , ¼ , ½ , ¾ , 1, , 51

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Measures 5
1000 m = 1 km How many metres is ½ km ? How many metres is ¼ km ? How many metres is ¾ km ? 52

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Measures 6
1 km 1000 m 250 m 250 m 250 m 250 m 250, 500, 750, 1000, , ¼ , ½ , ¾ , 1, , 53

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Measures 7
15, 30, , , ,
¼ , ½ , ¾ , , ,
min 3 6 54

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Multiplication
23 x 32
23 x 30 = 690
23 x 2 = 46
690
+46
736
46 x 23
46 x 20 = 920
46 x 3 = 138
920
+138
1058

57. Write each of these numbers as a product of prime factors.
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2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37…...
Prime Numbers
a number produced by multiplying
Product
Write each of these numbers as a product of prime factors.
Examples: 6= 2x3 and 15=3x5
14=
2x7 4=
2x2
12=
2x2x3

58. Write each of these numbers as a product of prime factors.
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Write each of these numbers as a product of prime factors. 10 9 22 21 33 25 49
= 2x5
= 3x3
= 2x11
= 3x7
= 3x11
= 5x5
= 7x7 20 18 66 42 99 50 98
= 2x2x5
= 2x3x3
= 2x3x11
= 2x3x7
= 3x3x11
= 2x5x5
= 2x7x7

59. Write each of these numbers as a product of prime factors.
100 90
220
210
330
250
490
2x2x5x5
3x3x2x5
2x11x2x5
3x7x2x5
3x11x2x5
5x5x2x5
7x7x2x5
400
180 72
360
900
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60. Find the prime factors of 60
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Find the prime factors of 60
Prime Numbers 2, 3, 5, 7, 11
10= 2x5
6= 2x3
60= 2x2x3x5 or 2² x 3 x 5
Find the prime factors of these numbers:
112 30 21 24
100
Find the prime factors of these numbers:
116 25 75 40 90

61. Recipe for 4 2 kg flour 2 litres water 500g sugar 1 kg strawberries
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Recipe for Splodge!
Recipe for 4
2 kg flour
2 litres water
500g sugar
1 kg strawberries
3 bags skittles
200g mushy peas
250g butter
Recipe for 8
4kg 4l
1000g or 1kg
2kg 6
400g
500g
Recipe for 6
3kg 3l
750g
1.5kg
4.5
300g
375g

62. Square Numbers or Perfect Squares
Cube Numbers
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1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
10x 10 =100
1²= 1x1= 1
2²= 2x2= 4
3²= 3x3= 9 4² 5² 6² 7² 8² 9²
10²
1³ = 1x1x1 =
2³ = 2x2x2 = 8
3³ = 3x3x =
4³ =
5³ = 5x5x5 =125
6³ =
7³ =
8³ =
9³ =
10³=10x10x10=1000
The area of this square is 400cm². (a) What is the length of one side?
(b) What is the perimeter of the square?
1 side = √400 = 20cm
perimeter = cm x 4 = 80cm

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73. Circumference = 2 x п x r = п d If r=5 C = 2 п r C = 2 x п x r
Circumference of Circles
Main Menu r
Circumference = 2 x п x r
= п d
If r=5
C = 2 п r
C = 2 x п x r
C = 2 x п x 5
C = 31.42
If r=3
C = 2 п r
C = 2 x п x r
C = 2 x п x 3
C = 18.85
If d=8
C = п d
C = п x d
C = п x 8
C =25.13

74. Area = п r² Area of Circles If r=3 If d=8 Area = п r² Area = п r²
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Area of Circles
Area = п r²
If r=3
Area = п r²
Area = п x 3 x3
Area = 28.27
= 28.3 (1dp)
If d=8
Area = п r²
Area = п x 4 x4
Area = 50.27
= 50.3 (3sf)

75. Area and Perimeter Perimeter = 2 ( 10 + 2 ) = 24m Area = 2 X 10 = 20m²
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10m 2m
Perimeter = 2 ( ) = 24m
Area = 2 X 10
= 20m²

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Area of Triangles h
height b
Find the area of a triangle with base 10cm and height 8 cm.
Area = (10 x 8) ÷ 2
Area = 80 ÷ 2
Area = 40 cm²
base
Area of = ½ ( b x h )
Area of = b x h 2

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